28
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Write a function which, given the first 12 digits of an ISBN-13 code, will calculate the entire ISBN via calculating and appending an appropriate check digit.

Your function's input is a string containing the first 12 digits of the ISBN. Its output is a string containing all 13 digits.

Formal specification

Write a function which, when given a string s consisting entirely of exactly 12 decimal digits (and no other characters), returns a string t with the following properties:

  • t consists of exactly 13 decimal digits (and no other characters);
  • s is a prefix of t;
  • the sum of all digits in odd positions in t (i.e. the first, third, fifth, etc.), plus three times the sum of all digits in even positions in t (i.e. the second, fourth, sixth, etc), is a multiple of 10.

Example / test case

Input
978030640615

Output
9780306406157

Victory condition

As a challenge, the shortest answer wins.

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  • 1
    \$\begingroup\$ Are input and output supposed to contain dashes or only the digits? \$\endgroup\$ – sepp2k Feb 1 '11 at 19:57
  • 1
    \$\begingroup\$ Updated the description, the input and output are only the digits \$\endgroup\$ – Kevin Brown Feb 1 '11 at 20:06
  • \$\begingroup\$ Is outputting the full ISBN-13 acceptable as well? \$\endgroup\$ – Mr. Llama Mar 2 '12 at 19:43
  • 6
    \$\begingroup\$ Note that questions should really be self contained, so it would be helpful to include a description of the algorithm here. \$\endgroup\$ – FlipTack Nov 19 '17 at 10:01
  • \$\begingroup\$ For me above post is poor of example test... There would be at last 10 isbn and one of that has to return 0 as last digit... \$\endgroup\$ – RosLuP Mar 9 at 19:34

45 Answers 45

14
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Golfscript - 25 chars

{...+(;2%+{+}*3-~10%`+}:f

Whole program version is only 19 chars

...+(;2%+{+}*3-~10%

Check back here for analysis later. Meanwhile check out my old uninspired answer

Golfscript - 32 chars

Similar to the luhn number calculation

{.{2+}%.(;2%{.+}%+{+}*~)10%`+}:f

Analysis for 978030640615

{...}:f this is how you define the function in golfscript
.       store an extra copy of the input string
        '978030640615' '978030640615'
{2+}%   add 2 to each ascii digit, so '0'=>50, I can get away with this instead
        of {15&}% because we are doing mod 10 math on it later
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55]
.       duplicate that list
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [59 57 58 50 53 50 56 54 50 56 51 55]
(;      trim the first element off
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [57 58 50 53 50 56 54 50 56 51 55]
2%      select every second element
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [57 50 50 54 56 55]
{.+}%   double each element by adding to itself
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [114 100 100 108 112 110]
+       join the two lists together
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55 114 100 100 108 112 110]
{+}*    add up the items in the list
        '978030640615' 1293
~       bitwise not
        '978030640615' -1294
)       add one
        '978030640615' -1293            
10%     mod 10
        '978030640615' 7
`       convert to str
        '978030640615' '7'
+       join the strings
        '9780306406157'
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  • \$\begingroup\$ After running the code through an interpreter, I think you can save some characters (in your 32 character Luhn-based solution) by getting rid of the first { and the last three characters; }:f. I wonder if the same thing can be done for the first solution... \$\endgroup\$ – Rob Oct 1 '12 at 11:45
  • \$\begingroup\$ @MikeDtrick, those characters are how GS defines a function. The 19 char version does what you suggest, but the question asked for a "function" \$\endgroup\$ – gnibbler Oct 2 '12 at 3:45
  • \$\begingroup\$ Oh okay, thank you for the explanation. \$\endgroup\$ – Rob Oct 2 '12 at 21:40
  • \$\begingroup\$ You don't need the :f (yeah, I know functions were commonly named back then). \$\endgroup\$ – Erik the Outgolfer Oct 22 '18 at 17:23
8
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Python - 44 chars

f=lambda s:s+`-sum(map(int,s+s[1::2]*2))%10`

Python - 53 chars

def f(s):d=map(int,s);return s+`-sum(d+d[1::2]*2)%10`
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  • \$\begingroup\$ I think f('9780306406159') outputs '97803064061598' instead of '9780306406157' \$\endgroup\$ – Eelvex Feb 2 '11 at 19:31
  • \$\begingroup\$ @Eelvex, the input string should always be 12 digits \$\endgroup\$ – gnibbler Feb 2 '11 at 20:47
  • \$\begingroup\$ Ah, somehow '9' sneaked in. Sorry... \$\endgroup\$ – Eelvex Feb 2 '11 at 21:19
7
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Haskell - 54 characters

i s=s++show(sum[-read[c]*m|c<-s|m<-cycle[1,3]]`mod`10)

This requires support for parallel list comprehensions, which is supported by GHC (with the -XParallelListComp flag) and Hugs (with the -98 flag).

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  • \$\begingroup\$ Do you not need to include that flag in the count? Besides that, you can replace [1,3] by [9,7] and remove the - which saves you a byte :) \$\endgroup\$ – ბიმო Nov 19 '17 at 22:02
7
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APL (27 characters)

F←{⍵,⍕10|10-(12⍴1 3)+.×⍎¨⍵}

I'm using Dyalog APL as my interpreter. Here's a quick explanation, mostly from right to left (within the function definition, F←{ ... }):

  • ⍎¨⍵: Execute/evaluate () each (¨) character given in the right argument ().
  • (12⍴1 3): Reshape () the vector 1 3 into a 12-element vector (repeating to fill the gaps).
  • +.×: Take the dot product (+.×) of its left argument ((12⍴1 3)) and its right argument (⍎¨⍵).
  • 10-: Subtract from 10.
  • 10|: Find the remainder after division by 10.
  • : Format the number (i.e., give a character representation).
  • ⍵,: Append (,) our calculated digit to the right argument.
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6
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PHP - 86 85 82 chars

function c($i){for($a=$s=0;$a<12;)$s+=$i[$a]*($a++%2?3:1);return$i.(10-$s%10)%10;}

Re-format and explanation:

function c($i){                     // function c, $i is the input

    for($a=$s=0;$a<12;)             // for loop x12 - both $a and $s equal 0
                                    // notice there is no incrementation and
                                    // no curly braces as there is just one
                                    // command to loop through

        $s+=$i[$a]*($a++%2?3:1);    // $s (sum) is being incremented by
                                    // $ath character of $i (auto-casted to
                                    // int) multiplied by 3 or 1, depending
                                    // wheter $a is even or not (%2 results
                                    // either 1 or 0, but 0 == FALSE)
                                    // $a is incremented here, using the
                                    // post-incrementation - which means that
                                    // it is incremented, but AFTER the value
                                    // is returned

    return$i.(10-$s%10)%10;         // returns $i with the check digit
                                    // attached - first it is %'d by 10,
                                    // then the result is subtracted from
                                    // 10 and finally %'d by 10 again (which
                                    // effectively just replaces 10 with 0)
                                    // % has higher priority than -, so there
                                    // are no parentheses around $s%10
}
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  • \$\begingroup\$ Exactly the approach I took in my C# answer. Seems PHP is ~9 characters more efficient! \$\endgroup\$ – Nellius Feb 2 '11 at 0:15
6
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Windows PowerShell, 57

filter i{$_+(990-($_-replace'(.)(.)','+$1+3*$2'|iex))%10}
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5
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Haskell, 78 71 66 characters

i s=s++(show$mod(2-sum(zipWith(*)(cycle[1,3])(map fromEnum s)))10)
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5
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Ruby - 73 65 chars

f=->s{s+((2-(s+s.gsub(/.(.)/,'\1')*2).bytes.inject(:+))%10).to_s}
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  • \$\begingroup\$ "\\1" -> '\1'? \$\endgroup\$ – Nemo157 Feb 1 '11 at 22:42
  • \$\begingroup\$ @Nemo, thanks, my ruby is a bit rusty \$\endgroup\$ – gnibbler Feb 1 '11 at 23:24
  • \$\begingroup\$ Use Ruby 1.9 syntax f=->s{...}. Save 6 chars. Also write s<<(...).to_sinstead of adding 48 and use Fixnum#chr. \$\endgroup\$ – Hauleth Mar 10 '12 at 16:58
4
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C# (94 characters)

string I(string i){int s=0,j=0;for(;j<12;)s+=(i[j]-48)*(j++%2<1?1:3);return i+((10-s%10)%10);}

With linebreaks/whitespace for readability:

string I(string i) 
{ 
    int s = 0, j = 0;
    for (; j < 12; )
        s += (i[j] - 48) * (j++ % 2 < 1 ? 1 : 3); 
    return i + ((10 - s % 10) % 10); 
}

Tested on several ISBNs from books on my shelf, so I know it's working!

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4
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Python - 91, 89

0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890
|         |         |         |         |         |         |         |         |         |
 def c(i):return i+`(10-(sum(int(x)*3for x in i[1::2])+sum(int(x)for x in i[::2]))%10)%10`
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  • \$\begingroup\$ Spaces are optional between the first argument and for (and in and the third) in a list comprehension so long as it can be split up by the parser (not using a variable name). -2 chars there. \$\endgroup\$ – Nick T Feb 3 '11 at 3:03
4
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Perl, 53 chars

sub i{$_=shift;s/(.)(.)/$s+=$1+$2*3/ge;$_.(10-$s)%10}
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4
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C# – 89 77 characters

string I(string s){return s+(9992-s.Sum(x=>x-0)-2*s.Where((x,i)=>i%2>0).Sum(x=>x-0))%10;}

Formatted for readability:

string I(string s)
{
    return s +
            (9992
            - s.Sum(x => x - 0)
            - 2 * s.Where((x, i) => i%2 > 0).Sum(x => x - 0)
            ) % 10;
}

We do not multiply by one or three, we just add everything, plus we add all even-placed characters one more time, multiplied by two.

9992 is large enough so that the sum of all ASCII characters is less than that (so that we may mod by 10 and be sure the result is positive, no need to mod by 10 twice), and is not divisible by zero because we add up all those extra 2*12*48 (twelve ASCII digits, weighed by 1 and 3) == 1152, which allows us to spare one extra character (instead of twice subtracting 48, we subtract 0 just to convert from char to int, but instead of 990, we need to write 9992).

But then again, even though much less beautiful ;-), this old-school solution gets us to 80 characters (but this is almost C-compatible):

string T(string i){int s=2,j=0;for(;j<12;)s+=i[j]*(9-j++%2*2);return i+s%10;}
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4
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J - 55 45 38

f=:3 :'y,":10|10-10|+/(12$1 3)*"."0 y'

eg

f '978030640615'
9780306406157

old way:

f=:,":@(10(10&|@-)(10&|@+/@((12$1 3)*(i.12)&(".@{))))
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  • 1
    \$\begingroup\$ (i.12)(".@{)y can be replaced with "."0 y \$\endgroup\$ – J Guy Oct 23 '18 at 6:36
3
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Ruby - 80 characters

def f s;s+(10-s.bytes.zip([1,3]*6).map{|j,k|(j-48)*k}.inject(:+)%10).to_s[0];end
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3
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dc, 44 chars

[d0r[I~3*rI~rsn++lndZ0<x]dsxx+I%Ir-I%rI*+]sI

Invoke as lIx, e.g:

dc -e'[d0r[I~3*rI~rsn++lndZ0<x]dsxx+I%Ir-I%rI*+]sI' -e '978030640615lIxp'
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3
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Q, 36 chars

{x,-3!10-mod[;10]sum(12#1 3)*"I"$'x}
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2
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D - 97 characters

auto f(string s){int n;foreach(i,c;s)n+=((i&1)*2+1)*(c-48);return s~cast(char)((10-n%10)%10+48);}

Formatted more legibly:

auto f(string s)
{
    int n;

    foreach(i, c; s)
        n += ((i & 1) * 2 + 1) * (c - 48);

    return s ~ cast(char)((10 - n % 10) % 10 + 48);
}

The verbosity of D's cast operator definitely makes it harder to write obsessively short code though.

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2
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Java - 161 Characters :(

int b[]=new int[a.length];
int d=0,n=0,j=1;
for(char c:a.toCharArray())b[d++]=Integer.valueOf(c+"");
for(int i:b)n+=(j++%2==0)?(i*3):(i*1);
return a+(10-(n%10));
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  • \$\begingroup\$ This answer does not satisfy the first requirement, since it is not a function. \$\endgroup\$ – han Oct 3 '12 at 17:49
1
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Q (44 chars)

f:{x,string 10-mod[;10]0+/sum@'2 cut"I"$/:x}
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  • \$\begingroup\$ This is actually incorrect I think \$\endgroup\$ – skeevey Mar 9 '12 at 19:53
1
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Scala 84

def b(i:String)=i+(10-((i.sliding(2,2).map(_.toInt).map(k=>k/10+k%10*3).sum)%10)%10)

Testing:

val isbn="978030640615"
b(isbn)

Result:

"9780306406157"
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1
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C, 80 79 characters

The function modifies the string in place, but returns the original string pointer to satisfy the problem requirements.

s;char*f(char*p){for(s=2;*p;s+=7**p++)s+=9**p++;*p++=48+s%10;*p=0;return p-13;}

Some explanation: Instead of subtracting 48 (the ASCII value of the digit 0) from each input character, the accumulator s is initialized so that it is modulo 10 equal to 48+3*48+48+3*48...+48+3*48 = 24*48 = 1152. The step 10-sum can be avoided by accumulating s by subtraction instead of addition. However, the module operator % in C would not give a usable result if s was negative, so instead of using s-= the multipliers 3 and 1 are replaced by -3=7 modulo 10 and -1=9 modulo 10, respectively.

Test harness:

#include <stdio.h>
#define N 12
int main()
{
     char b[N+2];
     fgets(b, N+1, stdin);
     puts(f(b));
     return 0;
}
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1
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Groovy 75, 66 chars

i={int i;it+(10-it.inject(0){t,c->t+(i++&1?:3)*(c as int)}%10)%10}

use:

String z = "978030640615"
println i(z)

-> 9780306406157
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1
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APL (25)

{⍵,⍕10-10|+/(⍎¨⍵)×12⍴1,3}
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  • \$\begingroup\$ The algo has to end with 10| because else it could return 10 in place of 0 \$\endgroup\$ – RosLuP Mar 9 at 19:31
1
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Perl 6, 29 bytes

{$_~-:1[.comb «*»(1,3)]%10}

Try it online!

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  • \$\begingroup\$ So base 1 can be used as a substitute for sum? Interesting! \$\endgroup\$ – Jo King Oct 23 '18 at 6:22
  • 2
    \$\begingroup\$ @JoKing It's actually a very old trick in the world of APL and J :) \$\endgroup\$ – Bubbler Oct 23 '18 at 6:31
  • \$\begingroup\$ Wrong result for 978186197371 it seems to be 8 not 9... \$\endgroup\$ – RosLuP Mar 9 at 17:11
  • \$\begingroup\$ Excuse me I think I had pasted the wrong number \$\endgroup\$ – RosLuP Mar 9 at 17:20
1
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Python 2, 78 76 bytes

lambda n:n+`10-(sum(int(a)+3*int(b)for a,b in zip(n[::2],n[1::2]))%10or 10)`

Try it online!

Takes a string as an argument.

Explanation:

Using python slice notation, converts a string into a list of character pairs. ("978030640615" -> [("9","7"), ("8", "0"), ("3", "0"), ("6", "4"), ("0", "6"), ("1", "5")] )

For that list of pairs, converts each item into an integer and returns a+3b.

Sums all the results.

Gets the sum modulo 10, OR 10 if the remainder is 0. (This prevents the final digit from being 10 instead of 0.)

Removes the remainder from 10 to get the check digit.

Converts the calculated check digit to a string via deprecated backtick expression.

Returns the original number plus the calculated check digit.

Edit:

Saved 2 byes by removing spaces (thanks Jo King!).

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  • \$\begingroup\$ You can remove the spaces before for and or \$\endgroup\$ – Jo King Oct 23 '18 at 6:19
  • \$\begingroup\$ Result to me that 978186197371 has last digit 8 and not 9... I get that number from the only link I print in my Apl solution \$\endgroup\$ – RosLuP Mar 9 at 17:09
  • \$\begingroup\$ Excuse me I think I have pasted the wrong number... \$\endgroup\$ – RosLuP Mar 9 at 17:18
1
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APL (Dyalog Unicode), 18 bytesSBCS

Anonymous tacit prefix function taking string as argument. Using Bubbler's approach.

⊢,∘⍕10|⍎¨+.×9 7⍴⍨≢

Try it online!

 length of argument (12)

9 7⍴⍨ cyclically reshape [9,7] to that length

+.× dot product of the following with that:

⍎¨ `evaluate each character

10| mod-10 of that

,∘⍕ prepend the following the the stringification of that:

 the unmodified argument

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1
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dc, 25 bytes

dn[A~9z^8+*rd0<M+]dsMxA%p

Try it online!

I know there's already a dc answer here, but 25<44 so I guess I feel 19 bytes of okay about it. This uses the fact that 8+9^z is equivalent to either -3 or -1 mod 10 depending on whether z is even or odd. So I use A~ to break the number into digits on the stack, but as I build the stack I multiply each digit by 8+9^z where z is the current stack size. Then I add them all as the function stack unrolls, and print the last digit.

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0
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MATLAB - 82 chars

function c(i)
[i num2str(mod(10-mod(sum(str2num(i(:)).*repmat([1;3],6,1)),10),10))]
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0
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R, 147 characters

f=function(v){s=as.numeric(strsplit(v,"")[[1]]);t=0;for(i in 1:12)if(i%%2==0)t=t+s[i]*3 else t=t+s[i];paste(v,(10-(t%%10))%%10,collapse="",sep="")}

Usage:

f("978030640615")
[1] "9780306406157"
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0
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J, 25

,[:":10|0(-+/)"."0*1 3$~#
   f =: ,[:":10|0(-+/)"."0*1 3$~#
   f '978030640615'
9780306406157
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